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 Miscellaneous math
 Posted 2016-07-19, 04:00 PM $\reverse \sum_{n=0}^{\infty} \frac{x^{4n}}{(4n)!} = \frac{\cos(x) + \cosh(x)}{2}$ $\reverse {n\choose k} = {n\choose n\cdot\sum\limits_{p=0}^{r-1} (-1)^p + k\cdot\prod\limits_{q=0}^{r-1} (-1)},\,r \in \mathbb{Z}$ "Stephen Wolfram is the creator of Mathematica and is widely regarded as the most important innovator in scientific and technical computing today." - Stephen Wolfram
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 Chruser

 Posted 2016-07-20, 12:56 PM in reply to Chruser's post "Miscellaneous math" Surely, there is a simpler method to select between 1 and 0 depending on odd/even than your summation?
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WetWired

 Posted 2016-07-20, 06:02 PM in reply to WetWired's post starting "Surely, there is a simpler method to..." The answer is 7
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-Spector-

 Posted 2016-07-20, 08:49 PM in reply to Chruser's post "Miscellaneous math" Wait! What an obtuse way to say that n choose k = n choose k or n choose n-k! Why not just $\reverse {n\choose k} = {n\choose n\cdot{(-1)^r+1\over2} + k\cdot(-1)^{r+1}},\,r \in \mathbb{Z}$ or even $\reverse {n\choose k} = {n\choose n-k}$ Last edited by WetWired; 2016-07-20 at 08:58 PM.
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WetWired

Posted 2016-07-25, 06:25 PM in reply to WetWired's post starting "Wait! What an obtuse way to say that n..."
WetWired said: [Goto]
 Wait! What an obtuse way to say that n choose k = n choose k or n choose n-k! Why not just $\reverse {n\choose k} = {n\choose n\cdot{(-1)^r+1\over2} + k\cdot(-1)^{r+1}},\,r \in \mathbb{Z}$ or even $\reverse {n\choose k} = {n\choose n-k}$
cuz u a bitch
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-Spector-

Posted 2016-07-31, 07:48 PM in reply to WetWired's post starting "Wait! What an obtuse way to say that n..."
WetWired said: [Goto]
 Wait! What an obtuse way to say that n choose k = n choose k or n choose n-k! Why not just $\reverse {n\choose k} = {n\choose n\cdot{(-1)^r+1\over2} + k\cdot(-1)^{r+1}},\,r \in \mathbb{Z}$ or even $\reverse {n\choose k} = {n\choose n-k}$

Yes, my "formula" is just the result of using the identity $\reverse {n\choose k} = {n\choose n-k}$ repeatedly, followed by a little algebra:

$\reverse {n\choose k} = {n\choose n-k} = {n\choose n-(n-k)} = {n\choose n-(n-(n-k))} = \dots$

I just thought the symmetry between the sum and product (and their upper limits) was fun.
"Stephen Wolfram is the creator of Mathematica and is widely regarded as the most important innovator in scientific and technical computing today." - Stephen Wolfram
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 Chruser

 Posted 2016-08-01, 06:25 PM in reply to Chruser's post starting "Yes, my "formula" is just the result of..." Since I'm just sitting here shit posting, what's your favorite "fun fact about math"? Mine is: X% of Y is equal to Y% of X
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-Spector-

 Posted 2016-08-02, 12:00 PM in reply to -Spector-'s post starting "Since I'm just sitting here shit..." $\reverse {x\over 100} \times y = {y\over 100} \times x$ $\reverse {x \times y \over 100} = {y \times x \over 100}$ $\reverse {x \times y} = {y \times x}$ $\reverse {x \times y} = {x \times y}$ $\reverse {x} = {x}$ $\reverse {1} = {1}$
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WetWired

 Posted 2016-08-03, 08:29 AM in reply to -Spector-'s post starting "Since I'm just sitting here shit..." If you had a wire as long as the circumference of the Earth and added just ten meters to it it would float 1.6 meters off the ground everywhere. If you could fold a piece of paper 42 times you would reach the moon and beyond. If you have 23 people in a room there's a 50% chance that two of them share a birthday. If you have 50 people in a room there's a 97% chance that two of them share a birthday. (1/2)+(1/4)+(1/8)+(1/16)+(1/32)+...=1 An infinite surface area can enclose a finite volume. Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" It is to your advantage to switch your choice. 10! seconds is exactly six weeks. If you properly shuffle a deck of cards, chances are the order in that deck has never been seen before. Given a solid ball in 3dimensional space, there exists a decomposition of the ball into a finite number of disjoint subsets, which can then be put back together in a different way to yield two identical copies of the original ball. This is known as the Banach-Tarski paradox.
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Demosthenes

Posted 2016-08-03, 11:44 AM in reply to Demosthenes's post starting "If you had a wire as long as the..."
Demosthenes said: [Goto]
 Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" It is to your advantage to switch your choice.
False. This is only true if he has to show you a door with a goat behind it regardless of your original choice.
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WetWired

 Posted 2016-08-12, 05:57 PM in reply to WetWired's post starting "False. This is only true if he has to..." WHat in the actual fuck... I can't even remember how to do fractions.
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Kazilla

Posted 2016-08-12, 06:04 PM in reply to Kazilla's post starting "WHat in the actual fuck... I can't even..."
Kazilla said: [Goto]
 WHat in the actual fuck... I can't even remember how to do fractions.

1/2 = .5 = 50%

now you know
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-Spector-

 Posted 2016-08-12, 08:35 PM in reply to -Spector-'s post starting "1/2 = .5 = 50% now you know" Yea but, 3/4 * 5/7 = X I don't even know where to begin. I think you have to find common denominator right? so 28. 21/28 * 20/28 right? Then you are supposed to flip one upside down? 21/28 * 28/20 then multiply them and then some division and you are good right? 100% not trolling btw, I legit forgot how to do that shit.
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Kazilla

 Posted 2016-08-13, 06:47 AM in reply to Kazilla's post starting "Yea but, 3/4 * 5/7 = X I don't even..." Er, no. when you multiply fractions you just multiply numerator * numerator and denominator * denominator so.. 3/4 * 5/7 = 15/28
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-Spector-

 Posted 2016-08-14, 09:08 PM in reply to Kazilla's post starting "Yea but, 3/4 * 5/7 = X I don't even..." You're thinking of addition.
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WetWired

 Posted 2016-08-15, 05:05 AM in reply to WetWired's post starting "You're thinking of addition." $\reverse \int_0^\infty \frac{\sin(x)}{x} \, \mathrm{d}x=\frac{\pi}{2}$ $\reverse \int_0^\infty \frac{\sin(x)}{x}\cdot\frac{\sin(x/3)}{x/3} \, \mathrm{d}x =\frac{\pi}{2}$ $\reverse \int_0^\infty \frac{\sin(x)}{x}\cdot\frac{\sin(x/3)}{x/3}\cdot\frac{\sin(x/5)}{x/5} \, \mathrm{d}x = \frac{\pi}{2}$ $\reverse \int_0^\infty \frac{\sin(x)}{x}\cdot\frac{\sin(x/3)}{x/3}\cdot\frac{\sin(x/5)}{x/5}\cdot\frac{\sin(x/7)}{x/7} \, \mathrm{d}x = \frac{\pi}{2}$ $\reverse \int_0^\infty \frac{\sin(x)}{x}\cdot\frac{\sin(x/3)}{x/3}\cdot\frac{\sin(x/5)}{x/5}\cdot\frac{\sin(x/7)}{x/7}\cdot\frac{\sin(x/9)}{x/9} \, \mathrm{d}x = \frac{\pi}{2}$ $\reverse \int_0^\infty \frac{\sin(x)}{x}\cdot\frac{\sin(x/3)}{x/3}\cdot\frac{\sin(x/5)}{x/5}\cdot\frac{\sin(x/7)}{x/7}\cdot\frac{\sin(x/9)}{x/9}\cdot\frac{\sin(x/11)}{x/11} \, \mathrm{d}x = \frac{\pi}{2}$ $\reverse \int_0^\infty \frac{\sin(x)}{x}\cdot\frac{\sin(x/3)}{x/3}\cdot\frac{\sin(x/5)}{x/5}\cdot\frac{\sin(x/7)}{x/7}\cdot\frac{\sin(x/9)}{x/9}\cdot\frac{\sin(x/11)}{x/11}\cdot\frac{\sin(x/13)}{x/13} \, \mathrm{d}x = \frac{\pi}{2}$ $\reverse \int_0^\infty \frac{\sin(x)}{x}\cdot\frac{\sin(x/3)}{x/3}\cdot\frac{\sin(x/5)}{x/5}\cdot\frac{\sin(x/7)}{x/7}\cdot\frac{\sin(x/9)}{x/9}\cdot\frac{\sin(x/11)}{x/11}\cdot\frac{\sin(x/13)}{x/13}\cdot\frac{\sin(x/15)}{x/15} \, \mathrm{d}x = \frac{467807924713440738696537864469}{935615849440 640907310521750000}~\pi$ "Stephen Wolfram is the creator of Mathematica and is widely regarded as the most important innovator in scientific and technical computing today." - Stephen Wolfram
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 Chruser

Posted 2016-09-04, 06:37 PM in reply to Chruser's post starting "\int_0^\infty \frac{\sin(x)}{x} \,..."
Chruser said: [Goto]
 $\reverse \int_0^\infty \frac{\sin(x)}{x} \, \mathrm{d}x=\frac{\pi}{2}$ $\reverse \int_0^\infty \frac{\sin(x)}{x}\cdot\frac{\sin(x/3)}{x/3} \, \mathrm{d}x =\frac{\pi}{2}$ $\reverse \int_0^\infty \frac{\sin(x)}{x}\cdot\frac{\sin(x/3)}{x/3}\cdot\frac{\sin(x/5)}{x/5} \, \mathrm{d}x = \frac{\pi}{2}$ $\reverse \int_0^\infty \frac{\sin(x)}{x}\cdot\frac{\sin(x/3)}{x/3}\cdot\frac{\sin(x/5)}{x/5}\cdot\frac{\sin(x/7)}{x/7} \, \mathrm{d}x = \frac{\pi}{2}$ $\reverse \int_0^\infty \frac{\sin(x)}{x}\cdot\frac{\sin(x/3)}{x/3}\cdot\frac{\sin(x/5)}{x/5}\cdot\frac{\sin(x/7)}{x/7}\cdot\frac{\sin(x/9)}{x/9} \, \mathrm{d}x = \frac{\pi}{2}$ $\reverse \int_0^\infty \frac{\sin(x)}{x}\cdot\frac{\sin(x/3)}{x/3}\cdot\frac{\sin(x/5)}{x/5}\cdot\frac{\sin(x/7)}{x/7}\cdot\frac{\sin(x/9)}{x/9}\cdot\frac{\sin(x/11)}{x/11} \, \mathrm{d}x = \frac{\pi}{2}$ $\reverse \int_0^\infty \frac{\sin(x)}{x}\cdot\frac{\sin(x/3)}{x/3}\cdot\frac{\sin(x/5)}{x/5}\cdot\frac{\sin(x/7)}{x/7}\cdot\frac{\sin(x/9)}{x/9}\cdot\frac{\sin(x/11)}{x/11}\cdot\frac{\sin(x/13)}{x/13} \, \mathrm{d}x = \frac{\pi}{2}$ $\reverse \int_0^\infty \frac{\sin(x)}{x}\cdot\frac{\sin(x/3)}{x/3}\cdot\frac{\sin(x/5)}{x/5}\cdot\frac{\sin(x/7)}{x/7}\cdot\frac{\sin(x/9)}{x/9}\cdot\frac{\sin(x/11)}{x/11}\cdot\frac{\sin(x/13)}{x/13}\cdot\frac{\sin(x/15)}{x/15} \, \mathrm{d}x = \frac{467807924713440738696537864469}{935615849440 640907310521750000}~\pi$
Is there a source, of proof of this? I have no idea how to do those integrals.
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Demosthenes

 Posted 2016-09-05, 08:05 AM in reply to Demosthenes's post starting "Is there a source, of proof of this? I..." He is the source and the proof, accept him.
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-Spector-

Posted 2016-09-06, 03:58 PM in reply to Demosthenes's post starting "Is there a source, of proof of this? I..."
Demosthenes said: [Goto]
 Is there a source, of proof of this? I have no idea how to do those integrals.

Borwein integral.
"Stephen Wolfram is the creator of Mathematica and is widely regarded as the most important innovator in scientific and technical computing today." - Stephen Wolfram
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 Chruser

Posted 2019-07-22, 11:07 AM in reply to Demosthenes's post starting "Is there a source, of proof of this? I..."
Demosthenes said: [Goto]
 Is there a source, of proof of this? I have no idea how to do those integrals.

Apparently there's a new, pretty cool physical interpretation of the Borwein integrals in terms of random walks:

Phys.org said:
 In the new paper, the physicists show that the movements of infinitely many random walkers can be used to model the emergence and disappearance of the patterns in the Borwein integrals. To begin, the random walkers all start at the point zero on the one-dimensional number line. For the first step, each walker is allowed to move a random distance of up to 1 unit, either left or right. For the second step, each walker may move a random distance of up to 1/3, then a random distance of up to 1/5, then 1/7, 1/9, etc. That is, each successive allowable step distance corresponds to the next value of the expression 1/(2n—1). The main question is, what is the fraction of random walkers at the starting point (the origin) after each time step? It turns out that the fraction (more precisely, the probability density) of walkers at the origin at each time step n corresponds to the solution to the Borwein integral using the same n value. As the physicists explain, for the first seven steps, the probability density that a walker ends up at the origin is always ½, which via the theorem above corresponds to an integral value of π. The key idea is that, up to this time, the density of walkers at the origin is the same as if the entire number line was uniformly populated with walkers. In reality, as the maximum distance of each step is restricted, only part of the number line is accessible, i.e., the walkers' world is finite. However, for the first seven steps, the walkers at the origin perceive that their world is infinite, since they do not possess any information about the existence of boundaries that would indicate that the world is finite. This is because none of those walkers that reached the outer boundary of their world (+1 or -1 after the first step) would have been able to make it back to the starting point in less than seven steps, even if taking the maximum size steps allowed and all in the direction toward the starting point. As these walkers had zero probability of showing up at the starting point before the eighth step, they could not affect the fraction of random walkers at the starting point. So for the first seven steps, the density of walkers at the origin is fixed at ½ (it is "protected"). But once those walkers that have reached +1 or -1 return to the origin, the situation changes. After the eighth step, it's possible that some of these walkers return to the starting point. Now these walkers act as "messengers" in the sense that their return to the starting point reveals the existence of a boundary, telling the other walkers at the origin that their world is finite, and therefore influencing the density of walkers at the origin.

Sources:

https://phys.org/news/2019-07-illusi...s-physics.html
https://arxiv.org/abs/1906.04545

So basically, if I understand it correctly, in the first step, you send out a "messenger" from the origin along the real line, which ends up at a random x coordinate between -1 and 1 (e.g. at 0.437737). Assume it ends up at x = 1 after one step. In the next step, this messenger moves up to 1/3 units (randomly) either left or right, e.g. between x = 1-1/3 = 2/3 and x = 1+1/3 = 4/3. Assume it ends up at x = 2/3. In the third step, it moves up to 1/5 units either left or right. Assume it ends up at x = 2/3 - 1/5 = 7/15 ≈ 0.46666...

Continuing in this fashion, and since

1 > 1/3
1 > 1/3 + 1/5
1 > 1/3 + 1/5 + 1/7
1 > 1/3 + 1/5 + 1/7 + 1/9
1 > 1/3 + 1/5 + 1/7 + 1/9 + 1/11
1 > 1/3 + 1/5 + 1/7 + 1/9 + 1/11 + 1/13

but

1 < 1/3 + 1/5 + 1/7 + 1/9 + 1/11 + 1/13 + 1/15,

what happens is effectively that a messenger that was initially sent out to x = 1 (or x = -1) is able to return to the origin after 8 steps to inform it (or rather, another messenger at the origin) that the "probability" of ending up at the origin is skewed away from what it "thinks" it is, since the walkers (unbeknownst to it) have restrictions on their step lengths.

Similar arguments based on causality or limits on information propagation speed can be used to evaluate a number of tricky integrals; see the arXiv paper.

As a side note, the above reminds me of a recent, small modification of the well-studied Pólya urn model that seems to describe how patterns in innovation (and Zipf's, Heaps' and Taylor's laws) arise: https://www.technologyreview.com/s/6...vations-arise/

It's pretty interesting how small (and quite intuitive, in retrospect, IMAO) changes to simple models can give rise to explanations that have eluded mathematicians for many years. It makes me think there's still a lot of room for messing around with formulas, axioms and ideas and looking at the resulting patterns numerically in order to make very significant discoveries.
"Stephen Wolfram is the creator of Mathematica and is widely regarded as the most important innovator in scientific and technical computing today." - Stephen Wolfram

Last edited by Chruser; 2019-07-22 at 11:37 AM.
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